Question

If a, b, c are in G.P., then
A. ${\text{a}}^{\text{2}}\text{,}{\text{b}}^{\text{2}}\text{,}{\text{c}}^{\text{2}}$ are in G.P.
B. ${\text{a}}^{\text{2}}\text{(b + c),}{\text{c}}^{\text{2}}\text{(a + b),}{\text{b}}^{\text{2}}\text{(a + c)}$ are in G.P.
C. ${\displaystyle \frac{\text{a}}{\text{b + c}}}\text{,}{\displaystyle \frac{\text{b}}{\text{c + a}}}\text{,}{\displaystyle \frac{\text{c}}{\text{a + b}}}$ are in G.P.
D. None of the above

Answer 1

Hint: We’ll first write the conditional equation for a, b, c, and then will note the value of b and c in terms of a and common ratio to create a relation in a, b, c and then will find the required answer with help of those equations.

Complete step by step answer:

Given data: a, b, c are in G.P.
From the given data i.e. a, b, c are in G.P., we can say that the common ratio will remain the same
$$\begin{gathered} {\displaystyle \frac{\text{b}}{\text{a}}}\text{ = }{\displaystyle \frac{\text{c}}{\text{b}}}..............\text{(i)} \\ \Rightarrow {\text{b}}^{\text{2}}\text{ = ac}.........\text{(ii)} \end{gathered}$$
Let the common ratio be ‘r’
$\therefore \text{b = ar}$
Squaring both sides of the above equation
$$\begin{gathered} {\text{b}}^{\text{2}}\text{ = }{\left(\text{ar}\right)}^{\text{2}} \\ \Rightarrow {\text{b}}^{\text{2}}\text{ = }{\text{a}}^{\text{2}}{\text{r}}^{\text{2}} \end{gathered}$$
Also,$\text{c = a}{\text{r}}^{\text{2}}$
Squaring both sides of the above equation
$$\begin{gathered} {\text{c}}^{\text{2}}\text{ = }{\left(\text{a}{\text{r}}^{\text{2}}\right)}^{\text{2}} \\ \Rightarrow {\text{c}}^{\text{2}}\text{ = }{\text{a}}^{\text{2}}{\text{r}}^{\text{4}} \end{gathered}$$
From the value of ${\text{a}}^{\text{2}}\text{,}{\text{b}}^{\text{2}}\text{,}{\text{c}}^{\text{2}}$ we can see that they are in G.P. with a common ratio of $\text{'}{\text{r}}^{\text{2}}\text{'}$
Common ratio=${\displaystyle \frac{{\text{b}}^{\text{2}}}{{\text{a}}^{\text{2}}}}\text{ = }{\displaystyle \frac{{\text{c}}^{\text{2}}}{{\text{b}}^{\text{2}}}}\text{ = }{\text{r}}^{\text{2}}$
Therefore, option(A) is correct

Note: We can also verify our solution with the help of an example let say 2,4,8 where
a=2
b=4
c=8
for option(A) ${\text{a}}^{\text{2}}\text{,}{\text{b}}^{\text{2}}\text{,}{\text{c}}^{\text{2}}$=$\text{4,16,64}$
since ${\displaystyle \frac{\text{16}}{\text{4}}}\text{ = }{\displaystyle \frac{\text{64}}{\text{16}}}\text{ = 4}$
therefore, ${\text{a}}^{\text{2}}\text{,}{\text{b}}^{\text{2}}\text{,}{\text{c}}^{\text{2}}$are in G.P.
(A)${\text{a}}^{\text{2}}\text{,}{\text{b}}^{\text{2}}\text{,}{\text{c}}^{\text{2}}$ are in G.P.
Therefore option (A) is correct
for option(B) ${\text{a}}^{\text{2}}\text{(b + c),}{\text{c}}^{\text{2}}\text{(a + b),}{\text{b}}^{\text{2}}\text{(a + c)}$=${\text{2}}^{\text{2}}\text{(4 + 8),}{\text{8}}^{\text{2}}\text{(2 + 4),}{\text{4}}^{\text{2}}\text{(2 + 8)}$
$\text{48,386,160}$
since ${\displaystyle \frac{\text{386}}{\text{48}}}\text{ = }{\displaystyle \frac{\text{193}}{\text{24}}}\ne {\displaystyle \frac{\text{160}}{\text{386}}}\text{ = }{\displaystyle \frac{\text{80}}{\text{193}}}$
therefore, ${\text{a}}^{\text{2}}\text{(b + c),}{\text{c}}^{\text{2}}\text{(a + b),}{\text{b}}^{\text{2}}\text{(a + c)}$are not in G.P.
for option(C) ${\displaystyle \frac{\text{a}}{\text{b + c}}}\text{,}{\displaystyle \frac{\text{b}}{\text{c + a}}}\text{,}{\displaystyle \frac{\text{c}}{\text{a + b}}}$=${\displaystyle \frac{\text{2}}{\text{4 + 8}}}\text{,}{\displaystyle \frac{\text{4}}{\text{8 + 2}}}\text{,}{\displaystyle \frac{\text{8}}{\text{2 + 4}}}$
${\displaystyle \frac{\text{1}}{\text{6}}}\text{,}{\displaystyle \frac{\text{2}}{\text{5}}}\text{,}{\displaystyle \frac{\text{4}}{\text{3}}}$
since ${\displaystyle \frac{{\displaystyle \frac{\text{2}}{\text{5}}}}{{\displaystyle \frac{\text{1}}{\text{6}}}}}\text{ = }{\displaystyle \frac{\text{12}}{\text{5}}}\ne {\displaystyle \frac{{\displaystyle \frac{\text{4}}{\text{3}}}}{{\displaystyle \frac{\text{2}}{\text{5}}}}}\text{ = }{\displaystyle \frac{\text{10}}{\text{3}}}$
therefore, ${\displaystyle \frac{\text{a}}{\text{b + c}}}\text{,}{\displaystyle \frac{\text{b}}{\text{c + a}}}\text{,}{\displaystyle \frac{\text{c}}{\text{a + b}}}$are not in G.P.